Finding perfect numbers (optimization)

Question

I coded up a program in C# to find perfect numbers within a certain range as part of a programming challenge . However, I realized it is very slow when calculating perfect numbers upwards of 10000. Are there any methods of optimization that exist for finding perfect numbers? My code is as follows:

using System;
using System.Collections.Generic;
using System.Linq;

namespace ConsoleTest
{
 class Program
 {
  public static List<int> FindDivisors(int inputNo)
  {
   List<int> Divisors = new List<int>();
   for (int i = 1; i<inputNo; i++)
   {
    if (inputNo%i==0)
     Divisors.Add(i);
   }
   return Divisors;
  }

  public static void Main(string[] args)
  { 
   const int limit = 100000;

   List<int> PerfectNumbers = new List<int>();
   List<int> Divisors=new List<int>();
   for (int i=1; i<limit; i++)
   {
    Divisors = FindDivisors(i);
    if (i==Divisors.Sum())
     PerfectNumbers.Add(i);
   }

   Console.Write("Output =");

   for (int i=0; i<PerfectNumbers.Count; i++)
   {
    Console.Write(" {0} ",PerfectNumbers[i]);
   }

   Console.Write("\n\n\nPress any key to continue . . . ");
   Console.ReadKey(true);
  }
 }
} 

Answer 1

Use the formula

testPerfect = 2n-1(2n - 1)

to generate possiblities then check wether the number is in fact perfect.

try this for some bedtime reading

Answer 2

Do perfect numbers change? No. Look here . Surely, they should be calculated once and then stored. In your case, the only results will be

6
28
496
8128

The next one is 33550336. Outside your range.

Answer 3

Just the obvious one from me: you don't need to check every divisor. No point looking for divisors past inputNo/2 . That cuts down half of the calculations, but this is not an order of magnitude faster.

Answer 4

解决此类问题的一种方法涉及在每个数字的内存中构建一个巨大的数组，然后将数字相除。

Answer 5

For anyone interested in a LINQ based approach, the following method worked quite well and efficiently for me in determining whether or not a caller supplied integer value is a perfect number.

bool IsPerfectNumber(int value)
{
    var isPerfect = false;

    int maxCheck = Convert.ToInt32(Math.Sqrt(value));
    int[] possibleDivisors = Enumerable.Range(1, maxCheck).ToArray();
    int[] properDivisors = possibleDivisors.Where(d => (value % d == 0)).Select(d => d).ToArray();
    int divisorsSum = properDivisors.Sum();

    if (IsPrime(divisorsSum))
    {
        int lastDivisor = properDivisors.Last();
        isPerfect = (value == (lastDivisor * divisorsSum));
    }

    return isPerfect;
}

For simplicity and clarity, my implementation for IsPrime(), which is used within IsPerfectNumber(), is omitted.

Answer 6

To continue from Charles Gargent's answer there is a very quick way to check if a Mersenne Number aka 2^n - 1 is prime. It is called the Lucas-Lehmer test The basic pseudocode though (taken from the Wikipedia page) is:

// Determine if Mp = 2p − 1 is prime for p > 2
Lucas–Lehmer(p)
    var s = 4
    var M = 2p − 1
    repeat p − 2 times:
        s = ((s × s) − 2) mod M
    if s == 0 return PRIME else return COMPOSITE

Answer 7

if your still looking for something to calculate perfect numbers. this goes through the first ten thousand pretty quick, but the 33 million number is a little slower.

public class Perfect {
private static Perfect INSTANCE = new Perfect();

public static Perfect getInstance() {
    return INSTANCE;
}

/**
 * the method that determines if a number is perfect;
 * 
 * @param n
 * @return
 */
public boolean isPerfect(long n) {
    long i = 0;
    long value = 0;
    while(++i<n){
        value = (0 == n%i?value+i:value);
    }
    return n==value;
}
}